3.1 Comment 1

Theorem 1.3 might seem obvious as one may think that

$$ \begin{align*} \lim_{n}f(W_{\tau_{n}})=f(W_{\tau_{\mathbb{D}}}) \end{align*} $$

is immediate. However, this is not obvious for the simple reason that $W_{\tau _{n}}$ does not approach the boundary of the unit disc radially! Theorem 1.4 naturally raises the converse question: if the PSEP is solvable for a given law $\mu $ , must the Hilbert transform $\mathcal {H}\{q({\vert \cdot \vert }/{\pi })\}\in L_{2\pi }^{1}$ ? We conjecture that the answer is yes.

3.2 Comment 2

Theorem 1.4 implies the existence of solutions for the PSEP when $\mu $ has a finite pth moment for some $p>1$ . This follows from (1.1).

A natural question is how to find functions with integrable Hilbert transforms in $L_{2\pi }^{1}$ . A naive way to do so is to compute the Hilbert transform and then check its integrability. However, this is hard in practice. The following result, due to Zygmund, is handy.

Theorem 3.1 (see [Reference Duren5, Ch. 4]).

Let u be a harmonic function in the unit disc such that

(3.1) $$ \begin{align} \sup_{0\leq r<1}\int_{0}^{2\pi}\vert u(re^{\theta i})\vert\!\log^{+}(\vert u(re^{\theta i})\vert)\,d\theta<+\infty. \end{align} $$

Then,

$$ \begin{align*} \sup_{0\leq r<1}\int_{0}^{2\pi}\vert H\{u\}(re^{\theta i})\vert \,d\theta<+\infty. \end{align*} $$

In particular, the condition (3.1) is, in some sense, the weakest that one can impose on u. Note that the $\log ^{+}(x)$ can be replaced by $\log (1+\vert x\vert )$ [Reference Garnett8]. A classical result in harmonic analysis says that if g is a convex function, then $g(u)$ is subharmonic. In particular, the integral in (3.1) is nondecreasing in r [Reference Rudin12]. So, in terms of the boundary function, the condition (3.1) can be replaced by

$$ \begin{align*} \int_{0}^{2\pi}\vert u(e^{\theta i})\vert\!\log^{+}(\vert u(e^{\theta i})\vert)\,d\theta<+\infty, \end{align*} $$

which was originally formulated. Based on this fact, the planar Skorokhod embedding problem becomes solvable once $\vert \varphi \vert \!\log ^{+}(\vert \varphi \vert )$ belongs to $L_{2\pi }^{1}$ .

Next, we provide an example where the distribution $\mu $ has only a first moment. Consider the distribution of the random variable

$$ \begin{align*} \xi=\frac{5}{(\ln(10)+1)^{2}}-\frac{10}{X(1-\ln({X}/{10}))^{3}} \end{align*} $$

with X uniformly distributed on $(0,1)$ . The underlying quantile function is

$$ \begin{align*} q(u)=\frac{5}{(\ln(10)+1)^{2}}-\frac{10}{u(1-\ln({u}/{10}))^{3}},\quad u\in(0,1). \end{align*} $$

Using Bertrand’s criterion for integrals, q belongs to $L^{p}$ for $p=1$ , but not for any higher p. Near zero, we have the approximation

$$ \begin{align*} \frac{\log^{+}({1}/{u(1-\ln(u))^{3}})}{u(1-\ln(u))^{3}}=-\frac{\log(u(1-\ln(u))^{3})}{u(1-\ln(u))^{3}}\approx-\frac{\log(u)}{u\vert\!\ln(u)\vert^{3}}-3\frac{\log(\vert\!\ln(u)\vert)}{u\vert\!\ln(u)\vert^{3}}. \end{align*} $$

Thus, the integral of $\vert q\vert \!\log ^{+}(\vert q\vert )$ is finite. Therefore, the Hilbert transform of $\varphi $ is in $L_{2\pi }^{1}$ (recall $\varphi (\theta )=q({\vert \theta \vert }/{\pi })$ ). Consequently, the Skorokhod embedding problem is solvable for this $\mu $ .

Remark 3.2. Gross’ construction of the map $G(z)$ is based on the calculation of the Fourier coefficients of $\varphi $ . Now, one can see that $G(z)$ can be simply expressed as

$$ \begin{align*} G(z)=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{e^{ti}+z}{e^{ti}-z}\varphi(t)\,dt. \end{align*} $$

Although this paper is primarily addressing the method introduced in [Reference Gross9], the approach is directly applicable, without modification, to the method described in [Reference Boudabra and Markowsky2].

3.3 Comment 3

Mathematicians tend to shy away from $L^{p}$ spaces when $p<1$ due to their unconventional properties, such as the failure of the triangle inequality, or, in relation to our context, the inability to construct Poisson integrals. However, we present an example that shows something might be done when $p<1$ . Using the formulae in [Reference Boudabra1, Reference Boudabra and Markowsky3], the c.d.f of the stopped planar Brownian motion upon leaving the Koebe domain $\mathcal {K}=\mathbb {C}-(-\infty ,-\tfrac 14]$ is given by

$$ \begin{align*} F(w)=1-{\frac{2}{\pi}}\arctan(\sqrt{-1-4w}) \end{align*} $$

for $w\in (-\infty ,-\tfrac 14]$ . That is,

$$ \begin{align*} \varphi(t)=F^{-1}({{\vert t\vert}/{\pi}})=-\frac{1}{4\sin({t}/{2})^{2}}. \end{align*} $$

Interestingly, the function $\varphi (t)$ coincides with the real part of the radial limit of the Koebe function $k(z)={z}/{(1-z)^{2}}$ . In [Reference Gardiner, Aikawa, Kumagai, Mizuta and Suzuki7], the author gathered some results about the existence of entire functions and harmonic functions with a predefined radial boundary function. The following natural question seems worth being addressed. How can we construct the univalent function whose real part of its radial limit matches $\varphi $ when the distribution $\mu $ has exclusively a finite pth moment for some $p<1$ ? Another related issue will be how to reconcile this with the fact that every univalent function belongs to $\boldsymbol {H}^{p}(\mathbb {D})$ for all $p<\tfrac 12$ ?

3.4 Comment 4

It is worth noting that Theorem 1.3 is still valid if f is a proper map or, more generally, a B-proper map. The notion of B-properness was introduced in [Reference Markowsky10]. Both types of functions ensure that if a sequence $(z_{n})_{n}$ in the input domain is approaching its boundary, then the image sequence $(f(z_{n}))_{n}$ is also approaching the boundary of the range of f.